Complex zeros and conjugate pairs

Warm-up

Three quick questions from earlier lessons.

Discover

Solve $x^2 = -9$ . Is there a real number whose square is $-9$ ?

No real number; the solutions are $3i$ and $-3i$
Yes: $3$ and $-3$

Key idea

The number $i$ has $i^2 = -1$ . Complex numbers $a + bi$ give every quadratic its zeros.

For

ax^2 + bx + c

: if

b^2 - 4ac < 0

, there are no real zeros but two complex zeros

\dfrac{-b \pm i\sqrt{4ac - b^2}}{2a}

. A polynomial with real coefficients has complex zeros in conjugate pairs:

p + qi

and

p - qi

.

Worked example

Solve $x^2 - 4x + 13 = 0$ .

b^2 - 4ac = 16 - 52 = -36

The discriminant decides.

x = \dfrac{4 \pm 6i}{2}

√(−36) = 6i.

x = 2 + 3i

or

x = 2 - 3i

Divide both parts by 2.

Your turn, guided

Your turn: $x^2 - 2x + 5 = 0$ .

b^2 - 4ac = -16

4 − 20.

x = 1 + 2i

or

x = 1 - 2i

√(−16) = 4i, and (2 + 4i) ÷ 2 = 1 + 2i. Type this first solution.

Practice ★

How many real zeros does $x^2 + 5x + 10$ have?
1. Two
2. One
3. None: two complex zeros
How many real zeros does $x^2 + 4x + 9$ have?
1. Two
2. One
3. None: two complex zeros
How many real zeros does $x^2 + 2x + 20$ have?
1. Two
2. One
3. None: two complex zeros

Watch out!

Practice ★★

Solve $x^2 - 2x + 5 = 0$ . Its solutions are $u + vi$ and $u - vi$ , with $v > 0$ .

Solve $x^2 - 8x + 25 = 0$ . Type each solution as $a + bi$ .
Solve $x^2 - 8x + 32 = 0$ . Type each solution as $a + bi$ .

A polynomial with real coefficients has the zero $4 + 4i$ . Which other number must be a zero?
1. $4 - 4i$
2. $-4 - 4i$
3. $-4 + 4i$
A polynomial with real coefficients has the zero $8 + 7i$ . Which other number must be a zero?
1. $8 - 7i$
2. $-8 - 7i$
3. $-8 + 7i$
A polynomial with real coefficients has the zero $4 + 5i$ . Which other number must be a zero?
1. $4 - 5i$
2. $-4 - 5i$
3. $-4 + 5i$

Exit ticket

Three questions to check you have got it.

Solve $x^2 - 8x + 25 = 0$ . Type each solution as $a + bi$ .
Solve $x^2 - 8x + 32 = 0$ . Type each solution as $a + bi$ .
Solve $x^2 - 4x + 5 = 0$ . Type each solution as $a + bi$ .

Summary card

Complex zeros

$i^2 = -1$ , so $\sqrt{-k^2} = ki$ .
$b^2 - 4ac < 0$ : no real zeros, two complex ones.
Real coefficients: $p + qi$ and $p - qi$ come together.

Watch out: the conjugate changes only the sign of the $i$ part.

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